Proof

If we succeed to prove the expression $p(m)$ for the base case natural number $m\in\mathbb N,$ then $p(m)$ becomes true.
Following the Peano axioms, $n+1$ is the successor of the natural number $n$.
If we succeed to prove the induction step $p(n)\Rightarrow p(n+1)$, then it means that given $p(n)$ is true for some $n\ge m$, then $p(n+1)$ is also true for the successor $n+1\ge m$.
By the same argument, it means that given $p(n+1)$ is true for some $n+1\ge m$, then $p(n+2)$ is also true for the successor $n+2\ge m$.
By the same argument, it means that given $p(n+2)$ is true for some $n+2\ge m$, then $p(n+3)$ is also true for the successor $n+3\ge m$.
By the same argument, ... etc..
It follows that $p(n)$ is true for all $n\ge m.$
By the definition of a valid logical argument, the proving principle by complete induction is valid, since the conclusion is true, if the (infintely many premisses, starting with the base case $m$ and following the induction steps $n\rightarrow n+1$ for all $n\ge m$ are all true.
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Hoffmann, Dirk W.: "Grenzen der Mathematik - Eine Reise durch die Kerngebiete der mathematischen Logik", Spektrum Akademischer Verlag, 2011